1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Center of Mass of Quarter-Circle Wire

  1. Aug 26, 2016 #1
    On page 252 of Kleppner's Introduction to Mechanics (2d. Ed.) in Example 7.5, Kleppner analyzes the forces and torques on a uniform rod that have been bent into a quarter-circle of radius R and length PI*R/2. His diagram is attached.

    Kleppner writes, "[t]he center of mass is halfway along the rod at PI*R/2.'' In the diagram, you see this is the point on the rod at its midway point (at polar coordinates (R, PI/4), where the weight vector W is shown).

    I'm not sure how he places the center of mass at this point. I calculated the center of mass as at polar coordinates (2^(3/2) R/PI, PI/4). I calculated it as follows. The mass per unit length is:
    \lambda = \frac{M}{\frac{2 \pi R}{4}} = \frac{2M}{\pi R}
    I calculated the center of mass as:
    \frac{1}{M} \int_C \boldsymbol{r} dm = \frac{1}{M} \int_0^{\pi/2} R (\boldsymbol{a_x} \cos \theta + \boldsymbol{a_y} \sin \theta ) \lambda R d\theta =
    \frac{2R}{\pi} \int_0^{\pi/2} (\boldsymbol{a_x} \cos \theta + \boldsymbol{a_y} \sin \theta) d\theta =
    \left. \frac{2R}{\pi} (\boldsymbol{a_x} \sin \theta - \boldsymbol{a_y} \cos \theta ) \right|_0^{\pi /2} =
    \frac{2R}{\pi} (\boldsymbol{a_x} + \boldsymbol{a_y}) = \boldsymbol{a_r} \frac{2 \sqrt{2} R}{\pi} + \boldsymbol{a_\theta} \frac{\pi}{4}
    This places the center of mass at (2^(3/2) R/PI, PI/4), or about 90% of the distance from the origin to the rod. Kleppner seems to say the center of mass is on the rod itself. Did I make a mistake?

    Many thanks for your help!

    Attached Files:

    Last edited: Aug 26, 2016
  2. jcsd
  3. Aug 26, 2016 #2
    I also get 2R/π for the x and y coordinates of the center of mass. Not sure about what you are trying to show with your approximation.
  4. Aug 26, 2016 #3
    Thanks, it's good to know I'm not missing something obvious here.

    I shouldn't have used the approximation symbol, I should've used the equals sign. I was just converting from Cartesian to polar.

    Again, thanks for checking my calculation.
  5. Jan 19, 2017 #4
    same confusion here in order to find the torque because of the weight we should consider weight to be acting about center of mass ( i.e center of gravity ) which is at (2r/pi , 2r/pi) but Kleppner has taken weight to be acting about the middle point on the rod which is on rod itself.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted